2011 AMC 10A Problem 13

P11-15~3 minPrint

Basic Counting Number Properties Split into exhaustive cases and handle each

How many even integers are there between 200 and 700 whose digits are all different and come from the set {1, 2, 5, 7, 8, 9}?

Show solution without answering

Key insight

The first digit must be 2 or 5 and the last digit must be 2 or 8. If the first digit is 2, the last digit can only be 8.

Solution

The numbers have three digits. Two of the places have strong rules, so fill those first.

Hundreds digit. The number is between $200$ and $700$ , so the hundreds digit is $2$ , $3$ , $4$ , $5$ or $6$ . Of these, only $2$ and $5$ are in the allowed set.
Units digit. The number is even, so the units digit is even. The allowed even digits are $2$ and $8$ .
The digits must all be different, so the choice of hundreds digit can affect the units digit. Split into two cases.

Case 1: hundreds digit $2$ . The units digit can't be $2$ again, so it must be $8$ . The tens digit can be any of the $4$ allowed digits not yet used ( $1, 5, 7, 9$ ). That gives $4$ numbers: $218, 258, 278, 298$ .

Case 2: hundreds digit $5$ . The units digit can be $2$ or $8$ : $2$ choices. For each, the tens digit is any of the $4$ allowed digits not yet used. That gives $2 \times 4 = 8$ numbers.

Total: $4 + 8 = 12$ .

The answer is $\boxed{\textbf{(A)}\ 12}$ .

Why this works

When counting numbers with digit rules, fill the most restricted places first (here the hundreds place for the range and the units place for being even). If one choice changes how many options another place has, split into cases instead of multiplying straight through.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The number is a three-digit integer. For it to lie between $200$ and $700$ , the hundreds digit must be $2$ or $5$ (the only allowed digits in that range). For it to be even, the units digit must be $2$ or $8$ .

Case 1: hundreds digit $2$ . The units digit cannot also be $2$ , so it is $8$ . The tens digit is any of the remaining $4$ digits. That gives $4$ numbers.

Case 2: hundreds digit $5$ . The units digit can be $2$ or $8$ ( $2$ choices), then the tens digit is any of the remaining $4$ digits. That gives $2 \cdot 4 = 8$ numbers.

Total: $4 + 8 = 12$ .

The answer is $\boxed{\textbf{(A)}\ 12}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Number Theory, Chapter 1: Integers: The Basics
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Algebra, Chapter 20: Special Functions (§20.3)